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NCERT Solutions · Class 9 Mathematics Math of Space: Surface Area and Volume

52 questions · 52 still being checked

Exercise Set 14.3 1–9 (part 3 of 6)

  1. For \(\displaystyle \pi\), use one of the following approximations: \(\displaystyle \pi \approx \frac{22}{7}\) or \(\displaystyle \pi \approx 3.14\)

    Exercise 1

    Find the total surface area of a cone, if its slant height is 21\displaystyle 21 m and diameter of its base is 24\displaystyle 24 m.

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    Radius is half the diameter.\[r = \frac{24}{2} = 12\ \text{m}, \qquad l = 21\ \text{m} \]\[\text{TSA} = \pi r l + \pi r^2 = \pi r (l + r) \]\[\text{TSA} = \frac{22}{7} \times 12 \times (21 + 12) \]\[\text{TSA} = \frac{22}{7} \times 12 \times 33 = \frac{8712}{7} \approx 1244.57 \]Answer: \(\displaystyle \approx 1244.57\ \text{m}^2\) (\(\displaystyle \pi = \tfrac{22}{7}\))

  2. Exercise 2

    Find the curved surface area of a right circular cone whose slant height is 10\displaystyle 10 cm and base radius is 7\displaystyle 7 cm.

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    \[\text{CSA} = \pi r l \]\[\text{CSA} = \frac{22}{7} \times 7 \times 10 = 220 \]Answer: \(\displaystyle 220\ \text{cm}^2\) (\(\displaystyle \pi = \tfrac{22}{7}\))

  3. Exercise 3

    The height of a cone is 16\displaystyle 16 cm, and its base radius is 12\displaystyle 12 cm. Find the curved surface area and the total surface area of the cone.

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    The height, radius and slant height form the right triangle in the figure.NCERT_Solution_Class9_Maths_Ch14_Ex14-3_Q3\[l^2 = h^2 + r^2 = 16^2 + 12^2 = 256 + 144 = 400 \Rightarrow l = 20 \]\[\text{CSA} = \pi r l = 3.14 \times 12 \times 20 = 753.6 \]\[\text{TSA} = \pi r (l + r) = 3.14 \times 12 \times (20 + 12) = 1205.76 \]Answer: CSA \(\displaystyle = 753.6\ \text{cm}^2\), TSA \(\displaystyle = 1205.76\ \text{cm}^2\) (\(\displaystyle \pi = 3.14\))

  4. Exercise 4

    A cone has a height of 15\displaystyle 15 cm. If its volume is 1570 cm3\displaystyle 1570 \mathrm{~cm}^3, find the radius of the base.

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    \[V = \frac{1}{3}\pi r^2 h \]\[1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15 = 15.7\, r^2 \]\[r^2 = \frac{1570}{15.7} = 100 \Rightarrow r = 10 \]Answer: \(\displaystyle 10\ \text{cm}\) (\(\displaystyle \pi = 3.14\))

  5. Exercise 5

    The curved surface area of a cone is 308 cm2\displaystyle 308 \mathrm{~cm}^2 and its slant height is 14\displaystyle 14 cm . Find
    (i)
    the radius of the base,
    (ii)
    the total surface area of the cone.

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    This solution has not been cross-checked against the answer printed in NCERT.

    (i) Use \(\displaystyle \text{CSA} = \pi r l \).\[308 = \frac{22}{7} \times r \times 14 = 44\,r \]\[r = \frac{308}{44} = 7 \](ii)\[\text{TSA} = \pi r (l + r) = \frac{22}{7} \times 7 \times (14 + 7) = 22 \times 21 = 462 \]Answer: (i) \(\displaystyle 7\ \text{cm}\); (ii) \(\displaystyle 462\ \text{cm}^2\) (\(\displaystyle \pi = \tfrac{22}{7}\))

  6. Exercise 6

    A joker's cap is in the form of a right circular cone of base radius 7\displaystyle 7 cm and height 24\displaystyle 24 cm. Find the area of the sheet required to make 10\displaystyle 10 such caps.

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    A cap is open at the base, so only the curved surface is needed.NCERT_Solution_Class9_Maths_Ch14_Ex14-3_Q6\[l^2 = 7^2 + 24^2 = 49 + 576 = 625 \Rightarrow l = 25 \]\[\text{CSA of one cap} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \]\[\text{sheet for 10 caps} = 10 \times 550 = 5500 \]Answer: \(\displaystyle 5500\ \text{cm}^2\) (\(\displaystyle \pi = \tfrac{22}{7}\))

  7. Exercise 7

    What length of tarpaulin 3\displaystyle 3 m wide is required to make a conical tent of height 8\displaystyle 8 m and base radius 6\displaystyle 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20\displaystyle 20 cm.

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    The tent has no floor, so the tarpaulin covers the curved surface only.NCERT_Solution_Class9_Maths_Ch14_Ex14-3_Q7\[l^2 = 8^2 + 6^2 = 100 \Rightarrow l = 10\ \text{m} \]\[\text{CSA} = \pi r l = 3.14 \times 6 \times 10 = 188.4\ \text{m}^2 \]\[\text{length} = \frac{\text{area}}{\text{width}} = \frac{188.4}{3} = 62.8\ \text{m} \]Add the $\displaystyle 20$ cm ($\displaystyle 0.2$ m) allowance.\[62.8 + 0.2 = 63 \]Answer: \(\displaystyle 63\ \text{m}\) (\(\displaystyle \pi = 3.14\))

  8. Exercise 8

    A right triangle with sides 6\displaystyle 6 cm, 8\displaystyle 8 cm and 10\displaystyle 10 cm is rotated through 360\displaystyle 360° about the side of 8\displaystyle 8 cm. Find the volume and the curved surface area of the solid so formed.

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    Check the triangle is right-angled, with the right angle between the $\displaystyle 6$ cm and $\displaystyle 8$ cm sides.\[6^2 + 8^2 = 36 + 64 = 100 = 10^2 \]Rotating about AB gives a cone (figure).NCERT_Solution_Class9_Maths_Ch14_Ex14-3_Q8\[h = AB = 8,\qquad r = BC = 6,\qquad l = AC = 10 \]\[V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 6^2 \times 8 = 301.44 \]\[\text{CSA} = \pi r l = 3.14 \times 6 \times 10 = 188.4 \]Answer: volume \(\displaystyle = 301.44\ \text{cm}^3\), CSA \(\displaystyle = 188.4\ \text{cm}^2\) (\(\displaystyle \pi = 3.14\); exactly \(\displaystyle 96\pi\) and \(\displaystyle 60\pi\))

  9. Exercise 9

    Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?

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    Let the cup have base radius \(\displaystyle R\) and height \(\displaystyle H\); the water surface has radius \(\displaystyle r\) at depth \(\displaystyle \tfrac{H}{2}\). NCERT_Solution_Class9_Maths_Ch14_Ex14-3_Q9 In the axial section \(\displaystyle PM \parallel AO\), so \(\displaystyle \triangle VPM \sim \triangle VAO\). \[\frac{r}{R} = \frac{VM}{VO} = \frac{H/2}{H} = \frac12 \quad \Rightarrow \quad r = \frac{R}{2} \] The water is itself a cone of radius \(\displaystyle r\) and height \(\displaystyle \tfrac{H}{2}\). \[V_{\text{water}} = \frac13 \pi r^2 h = \frac13 \pi \left(\frac{R}{2}\right)^2 \left(\frac{H}{2}\right) = \frac18 \cdot \frac13 \pi R^2 H \] \[V_{\text{cup}} = \frac13 \pi R^2 H \] \[\frac{V_{\text{water}}}{V_{\text{cup}}} = \frac18 \] Answer: \(\displaystyle \dfrac18\) of the cup's volume.